The remainder (or modulus) is the amount "left over" after performing the division of two integers which do not divide evenly, that is, where the result of the division cannot be expressed as an integer.

## Remainder theorem

It states that the remainder of the division of a polynomial *f(x)* by a linear polynomial *x-a* is equal to *f(a)*.

**Proof**

The polynomial remainder theorem follows from the definition of polynomial long division; denoting the divisor, quotient and remainder by, respectively, g(x)g(x), q(x)q(x), and r(x)r(x), polynomial long division gives a solution of the equation

f(x)=q(x)g(x)+r(x), where the degree of r(x) is less than that of g(x).

If we take g(x)=x-a as the divisor, giving the degree of r(x) as 0, i.e. r(x)=r

f(x)=q(x)(x−a)+r.

Setting x=a we obtain

f(a)=r.

## Remainder Calculation

A number S = N1 + N2 + N3... when divided by d gives the remainder R = r1 + r2 + r3... where r1 = remainder of N1 divided by d, r2 = remainder of N2 divided by d and so on. This applies to all arithmetic operations including addition(as shown), subtraction, multiplication, and division. Let us consider the following question:

Find the remainder when 18 x 27 is divided by 13. Now, 18 x 27 = (13 + 5) x (26 + 1) So the remainder of 18, when divided by 13, will be 5, and the remainder of 27, when divided by 13, will be 1. So, remainder = (5 × 1)/13 Therefore, remainder = 5. This is the remainder when 18 x 27 is divided by 13.

## Special Cases in remainder calculation

**When both the dividend and the divisor have a factor in common**

Let N be a number and Q and R be the quotient and the remainder when N is divided by the divisor D.

Hence N = Q x D + R

Let N = k x A and D = k x B where k is the HCF of N and D and k > 1. Hence kA = Q x kB + R.

Let Q1 and R1 be the quotient and the remainder when A is divided by B. Hence A = B x Q1 + R1. Putting the value of A in the previous equation and comparing we get.

k(B x Q1 + R1) = Q x kb + R

R = kR1

Hence to find the remainder when both the dividend and the divisor have a factor in common,

- Take out the common factor (i.e. divide the numbers by the common factor)
- Divide the resulting dividend(A) by resulting divisor(B) and find the remainder(R1)
- The real remainder R is this remainder R1 multiplied by the common factor (k)

**Concept of negative remainders**

Consider a case when 13 x 15 is divided by 7. Using the normal method: (13 × 15)/7 → (6 × 1)/7 → 6/7 → 6 Now using concept of negative remainders: (13 × 15)/7 → (-1 × 1)/7 → (-1)/7 → -1 The negative remainder is -1. Whenever we have a negative remainder, we subtract it from the divisor. Thus, it becomes (7 - 1) = 6. The remainder, therefore, is 6. The concept of negative remainder makes our work simpler.

## Euler's theorem

If M and N are two numbers coprime to each other, i.e. HCF(M,N)=1 and N=(a^{p})(b^{q})(c^{r})..., Remainder of M^{φ(N)} when divided by N, is *1*.

and is known as the Euler's Totient function( It is also the number of numbers less than and prime to N.)

## Fermat's theorem

If P is a prime number and N is prime to P, then (N^P) - N is divisible by P.

**Fermat's little theorem**

If N in the above Euler's theorem is a prime number, then

Therefore, if M and N are coprime to each other and N is a prime number, Remainder of M^(N-1) when divided by N = 1